Question

Two tuning forks A and B of frequency 512 Hz are sounded together produce 5 beats/s. If the fork A is thin loaded with a piece of wax and found that beats occur at shorter intervals. Then the natural frequency of A will be
A. 517
B. 507
C. 512
D. 510

Answer 1

Correct Answer - B
`n_(A) = n_(B) +- 5`
`= 512 +- 5 = 517 or 507 Hz`
When `n_(A)` is loaded with little wax and sounded with `n_(B)` it produce beats of shorter time interval, that means beat frequency increases
`n_(A) = 512 +- 6`
`= 518 or 506`
`n_(A) = 507 Hz`